 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2019/iii/paper-202/5/2/solution

Since

B_{t} = \int_{0}^{t} 1 d B_{s}

E [B_{t} \int_{0}^{t} f (s) d B_{s}] = \int_{0}^{t} 1 \cdot f (s) d s = \int_{0}^{t} f (s) d s .

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